Bounds for the number of real roots

Worked illustration 1 (bound attained exactly). $P(x)=(x+1)(x-1)(x-2)(x-i)(x+i)=x^5-2x^4-x+2$ has roots $-1,1,2,i,-i$. $P(x)$'s coefficient signs (nonzero ones) are $+,-,-,+$: 2 sign changes, so at most $2$ positive roots. $P(-x)=-x^5-2x^4+x+2$ has signs $-,-,+,+$: 1 sign change, so at most $1$ negative root. The actual roots $1,2$ (positive, count $=2$) and $-1$ (negative, count $=1$) show both bounds are attained exactly here — $i,-i$ are of course neither positive nor negative.

Worked illustration 2 (bound not attained — a weaker, still useful, conclusion). $(x+2)(x+3)(x-i)(x+i)=x^4+5x^3+7x^2+5x+6$ has all-positive coefficient signs: 0 sign changes, so 0 positive roots — a sharp, fully-attained conclusion here (there genuinely are none). $P(-x)=x^4-5x^3+7x^2-5x+6$ has signs $+,-,+,-,+$: 4 sign changes, so at most $4$ negative roots — but the actual roots are only $-2,-3$ (just $2$ negative roots): the bound of $4$ is not attained, only the weaker even-difference guarantee ($4-2=2$, even ✓) holds. …